Question

agerg

Answer 1

@sourwing @phi @RadEn

Answer 2

r is the radius of the semi circle on top and x is the length height of the rectangle part

Answer 3

i have to be honest @habibmatatta i have no clue

Answer 4

no worries

Answer 5

thanks for understanding

Answer 6

is this calculus ?

Answer 7

precalc

Answer 8

the first thing you need is an equation for the perimeter
can you write one down?

Answer 9

is it 2x + (2 + pi)(r)?

Answer 10

yes,
P = 2x + 2r + pi r

the area of the rectangle is 2r * x
the area of the ½ circle is ½ pi r^2
total area A= 2rx + ½ pi r^2

We could at this point solve
P = 2x + 2r + pi r
for x (remember P is a constant)
and sub that value into the second equation for Area.

Can you do that ?

Answer 11

i have already solved it for r. i got r=(P-2x)/(2+pi)

Answer 12

oh im still doin alegebra one

Answer 13

I would solve for x.. if you use r, then you have to square r to use it ... it looks harder

Answer 14

ok x=(P-r(2+pi))/2

Answer 15

now use that for x in

A= 2rx + ½ pi r^2

Answer 16

ok A=2((P-r(2+pi))/2)r + ½ pi r^2

Answer 17

can you simplify that ?

Answer 18

yes A=r(P-r(2+pi))+(pi)r^2/2

Answer 19

keep going...

Answer 20

can't go farther than that

Answer 21

the variable is "r". all the other things are constants. You should put the expression in standard form: coefficient * r to highest exponent + coeff2 * r ^ next lower exponent and so on...

Answer 22

so (pi)r^2+2((P-r(2+pi))/2)r

Answer 23

start with

A=2((P-r(2+pi))/2)r + ½ pi r^2

I would simplify the first term, by "canceling" the 2
A=(P-r(2+pi))r + ½ pi r^2

now distribute the r (inside the parens
A= (P- 2r - pi r) r + ½ pi r^2
now distribute the r again
A= Pr - 2 r^2 - pi r^2 + ½ pi r^2

now collect terms
can you finish ?

Answer 24

(-1/2(pi)-2)r^2+Pr?

Answer 25

yes,
we can factor a -1 out, and write it as

-( 2+ pi/2) r^2 + P r
or , changing 2 to 4/2 so we can add the two terms in the parens

A= - ( (4+pi)/2 ) r^2 + Pr

that is the equation of a parabola (i.e. y = -a x^2 + bx)
it has a maximum at its vertex.
can you find the "x" value ( r value in this case) of the vertex... use x value of vertex = -b/(2a)

Answer 26

This shows how to find the vertex
http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php
notice that because the "a" coefficient in this problem is negative, the parabola is "upside down" and has a max (not a min)

Answer 27

yup i got it. thanks!!